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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: dmk007 on July 07, 2008, 08:38:12 AM

Title: % of nitrite ion??
Post by: dmk007 on July 07, 2008, 08:38:12 AM

I'd appreciate if someone could post a solution to this

Balance the following equations

Ce(SO4)2 + NaNO2 + H20 - Ce2(SO4)3 + NaNO3 + H2SO4

(NH4)2SO4FeSO4 + Ce(SO4)2 - Fe2(SO4)3 + Ce2(SO4)3 + (NH4)2SO4

In a laboratory experiment to calculate the % of nitrite ion NO2 in a salt 1.639 g of sodium nitrogen (IV) oxide, NaNO2 were made up to 500cm3 with deionised water. 25cm3 of this solution were added to 50cm3 of cerium (IV) sulphate
Ce(SO4)2.4H20 and 10cm3 of 2M sulphuric acid in a conical flask. The concentration of the cerium (IV) sulphate was 39.093 gdm-3. After 5 minutes the excess cerium (IV) sulphate was titrated with a standard solution of ammonium iron (II) sulphate hexahydrate, (NH4)2SO4FeSO4.6H20. If this required 23.7cm3 of iron salt and if the concentration of the iron salt was 41.196gdm-3 calculate the % of nitrite ion in the sodium nitrogen (IV) oxide salt.

Title: Re: % of nitrite ion??
Post by: Astrokel on July 07, 2008, 10:11:57 AM

hey dmk,

Did you come up with the equations yourself? Ce(SO4)2.4H20, where is the water account for?

Title: Re: % of nitrite ion??
Post by: dmk007 on July 07, 2008, 12:04:28 PM

well thats what my examiner had written do you know what should it be?

Title: Re: % of nitrite ion??
Post by: Astrokel on July 07, 2008, 01:17:56 PM

Sorry about my previous post,

Balance your equations first.

You have to find out the actual mole of Ce(SO4)2 actually reacted with NaNO2 first = (actual mole of Ce(SO4)2 used) - (mole of Ce(SO4)2 reacted with the iron salt)

Afterward, its easy to obtain the %.